Chapter 1 Topics in Analytic Geometry

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MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 44Orthogonal ProjectionsThevectorcomponentsofv alonge 1 ande 2 in(3.8)arealsocalledtheorthogonal projectionsof v on e 1 and e 2 and are denoted byproj e1v = (v·e 1 )e 1 and proj e2v = (v·e 2 )e 2In general, if e is a unit vector, then we define the orthogonal projection of v on e tobeproj e v = (v·e)e (3.10)The orthogonal projection of v on an arbitrary nonzero vector b can be obtained by normalizingb and then applying Formula (3.10); that is,( )( )b bproj b v = v·‖b‖ ‖b‖which can be rewritten asproj b v = v·b‖b‖ 2 b (3.11)Moreover, if we subtract proj b v from v, then the resulting vectorv−proj b vwill be orthogonal to b; we call this the vector component of v orthogonal to b.v−proj b vvvv−proj b vbproj b vproj b vbAcute angle between v and bObtuse angle between v and bExample 3.18 Find the orthogonal projection of v = i +j+k on b = 2i+2j, and thenfind the vector component of v orthogonal to b.Solution .........WorkRecall that we define the work W done on the object by a constant force of magnitude Facting in the direction of motion over the distance d to beW = Fd = force × distance (3.12)If we let F denote a force vector of magnitude ‖F‖ = F acting in the direction of motion,then we can write (3.12) asW = ‖F‖d

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 45Moreover, if we assume that the object moves along a line from point P to point Q, thend = ‖ −→ PQ‖, so that the work can be expressed entirely in vector form asW = ‖F‖‖ −→ PQ‖The vector −→ PQ is called the displacement vector for the object.In the case where a constant force F is not in the direction of motion, but rather makesan angle θ with the displacement vector, then we define the work W done by F to beW = (‖F‖cosθ)‖ −→ PQ‖ = F·−→ PQ (3.13)•P‖F‖FWork = ‖F‖‖ −→ PQ‖•Q•P‖F‖Fθ‖F‖cosθWork = (‖F‖cosθ)‖ −→ PQ‖•QExample 3.19 A force F = 8i + 5j in pound moves an object from P(1,0) to Q(7,1),distance measured in feet. How much work is done?Solution .........Example 3.20 A wagon is pulled horizontally by exerting a constant force of 10lb on thehandle at an angle of 60 ◦ with the horizontal. How much work is done in moving the wagon50 ft?Solution .........3.4 Cross ProductDeterminantsBefore we define the cross product, we need to define the notion of determinant.Definition 3.4 The determinant of a 2×2 matrix of real number is defined by∣ a ∣1 a 2∣∣∣= ab 1 b 1 b 2 −a 2 b 1 .2Definition 3.5 The determinant of a 3×3 matrix of real number is defined as a combinationof three 2×2 determinants, as follows:∣ a 1 a 2 a 3∣∣∣∣∣∣ ∣ ∣ ∣ ∣ ∣ ∣∣∣ b b 1 b 2 b 3 = a 2 b 3∣∣∣∣∣∣ b1 −a 1 b 3∣∣∣∣∣∣ b∣ cc 1 c 2 c 2 c 2 +a 1 b 2∣∣∣3 c 1 c 3 (3.14)3 c 1 c 23Note that Equation (3.14) is referred to as an expansion of the determinant alongthe first row.

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